remarks on some of the smarandache’s problem. part 2

8/12/2019 Remarks on some of the Smarandaches problem. Part 2

1/26

Scientia MagnaVol. 1 (2005), No. 2, 1-26

Remarks on some of the Smarandachesproblem. Part 2

Mladen V. Vassilev , Missana and Krassimir T. Atanassov 5,V.Hugo Str., Soa-1124, Bulgaria

e-mail:[email protected] CLBME-Bulg. Academy of Sci., P.O.Box 12, Soa-1113, Bulgaria,

e-mail:[email protected]

To Dr. Florentin Smarandachefor his 50th birthday

0. In 1999, the second author of this remarks published a book over 30 of Smarandachesproblems in area of elementary number theory (see [1, 2]). After this, we worked over new 20problems that we collected in our book [28]. These books contain Smarandaches problems,described in [10, 16]. The present paper contains some of the results from [28].

In [16] Florentin Smarandache formulated 105 unsolved problems, while in [10] C.Du-mitresu and V. Seleacu formulated 140 unsolved problems of his. The second book containsalmost all the problems from [16], but now each problem has unique number and by this reasonin [1, 28] and here the authors use the numeration of the problems from [10].

In the text below the following notations are used.

N - the set of all natural numbers (i.e., the set of all positive integers);[x] - oor function (or also so called bracket function) - the greatest integer which is notgreater than the real non-negative number x; - Riemanns Zeta-function; - Eulers Gamma-function; - the prime counting function, i.e., (n ) denotes the number of prime p such that p n ;]x[ - the largest natural number strongly smaller than the real (positive) number x ;x - the inferior integer part of x, i.e, the smallest integer greater than or equal to x.

For an arbitrary increasing sequence of natural number C {cn }n =1 we denote by C (n )the number of terms of C , which are not greater than n . When n < c 1 we put C (n ) = 0.

1. The results in this section are taken from [8].The second problem from [10] (see also 16-th problem from [16]) is the following:Smarandache circular sequence:

1

1, 12, 21

2, 123, 231, 312

3, 1234, 2341, 3412, 4123

4,

12345, 23451, 34512, 45123, 51234

5,


2/26

2 Mladen V. Vassilev, Missana and Krassimir T. Atanassov No. 2

123456, 234561, 345612, 456123, 561234, 612345

6,

Let f (n ) be the n -th member of the above sequence. We shall prove the following

Theorem 1.1. For each natural number n:

f (n ) = s (s + 1) . . . k 12 . . . (s 1),where

k k(n ) =] 8n + 1 1

2 [

ands s (n ) = n

k(k + 1)2

.

2. The results in this section are taken from [25].The eight problem from [10] (see also 16-th problem from [16]) is the following:Smarandache mobile periodicals (I):

. . . 0 0 0 0 0 0 1 0 0 0 0 0 0. . .

. . . 0 0 0 0 0 1 1 1 0 0 0 0 0. . .

. . . 0 0 0 0 1 1 0 1 1 0 0 0 0. . .

. . . 0 0 0 0 0 1 1 1 0 0 0 0 0. . .

. . . 0 0 0 0 0 0 1 0 0 0 0 0 0. . .

. . . 0 0 0 0 0 1 1 1 0 0 0 0 0. . .

. . . 0 0 0 0 1 1 0 1 1 0 0 0 0. . .

. . . 0 0 0 1 1 0 0 0 1 1 0 0 0. . .

. . . 0 0 0 0 1 1 0 1 1 0 0 0 0. . .

. . . 0 0 0 0 0 1 1 1 0 0 0 0 0. . .

. . . 0 0 0 0 0 0 1 0 0 0 0 0 0. . .

. . . 0 0 0 0 0 1 1 1 0 0 0 0 0. . .

. . . 0 0 0 0 1 1 0 1 1 0 0 0 0. . .

. . . 0 0 0 1 1 0 0 0 1 1 0 0 0. . .

. . . 0 0 1 1 0 0 0 0 0 1 1 0 0. . .

. . . 0 0 0 1 1 0 0 0 1 1 0 0 0. . .

. . . 0 0 0 0 1 1 0 1 1 0 0 0 0. . .

. . . 0 0 0 0 0 1 1 1 0 0 0 0 0. . .

. . . 0 0 0 0 0 0 1 0 0 0 0 0 0. . .


3/26

Vol. 1 Remarks on Some of the Smarandaches Problem. Part 2 3

. . . 0 0 0 0 0 0 1 0 0 0 0 0 0. . .

. . . 0 0 0 0 0 1 1 1 0 0 0 0 0. . .

. . . 0 0 0 0 1 1 0 1 1 0 0 0 0. . .

. . . 0 0 0 1 1 0 0 0 1 1 0 0 0. . .

. . . 0 0 1 1 0 0 0 0 0 1 1 0 0. . .

. . . 0 1 1 0 0 0 0 0 0 0 1 1 0. . .

. . . 0 0 1 1 0 0 0 0 0 1 1 0 0. . .

This sequence has the form

1, 111, 11011, 111 , 1

3, 1, 111, 11011, 1100011, 11011, 111, 1

7,

1, 111, 11011, 1100011, 110000011, 1100011, 11011, 111, 1

9, . . .

All digits from the above table generate an innite matrix A. We described the elementsof A.

Let us take a Cartesian coordinate system C with origin in the point containing element1 in the topmost (i.e., the rst) row of A. We assume that this row belongs to the ordinateaxis of C (see Fig. 1) and that the points to the right of the origin have positive ordinates.

The above digits generate an innite sequence of squares, located in the half-plane (de-termined by C ) where the abscissa of the points are nonnegative. Their diameters have theform

110 . . . 011 .

Exactly one of the diameters of each of considered square lies on the abscissa of C . It canbe seen (and proved, e.g.,by induction) that the s-th square, denoted by Gs (s = 0 , 1, 2, . . . )has a diameter with length 2 s + 4 and the same square has a highest vertex with coordinatess 2 + 3 s, 0 in C and a lowest vertex with coordinates s2 + 5 s + 4 , 0 in C .

Let us denote by a k,i an element of A with coordinates k, i in C .First, we determine the minimal nonnegative s for which the inequality

s2 + 5 s + 4 kholds. We denote it by s (k). Directly it is seen the followingLemma 2.1 The number s(k) admits the explicit representation:

s(k) =

0, if 0 k 4 4k +9 52 , if k 5 and 4k + 9 isa square of an integer

4k +9 52 + 1 , if k 5 and 4k + 9 isnot a square of an integer

(2.1)


4/26


and the inequality(s(k)) 2 + 3 s(k) k (s(k)) 2 + 5 s(k) + 4 (2 .2)

hold.Second, we introduce the integer (k) and (k) by

(k) k (s(k)) 2 3s(k), (2.3)(k) (s(k)) 2 + 5 s(k) + 4 k. (2.4)

From (2.2) we have (k) 0 and (k) 0. Let P k be the innite strip orthogonal to theabscissa of C and lying between the straight lines passing through those vertices of the squareG s (k ) lying on the abscissa of C . Then (k) and (k) characterize the location of point withcoordinate k, i in C in strip P k . Namely, the following assertion is true.Proposition 2.1. The elements ak,i of the innite matrix A are described as follows:if k (s(k)) 2 + 4 s(k) + 2, then

a k,i =0, if (k) < |i| or (k) |i|+ 2 ,1, if |i| (k) |i|+ 1

if k (s(k)) 2 + 4 s(k) + 2, then

a k,i =0, if (k) < |i| or (k) |i|+ 2 ,1, if |i| (k) |i|+ 1

where here and below s (k) is given by (2.1), (k) and (k) are given by (2.3) and (2.4), respec-tively.

Below, we propose another description of elements of A, which can be proved (e.g., byinduction) using the same considerations.

a k,i =

1, if k, i {(s (k)) 2 + 3 s (k), 0 , (s(k))2 + 5 s(k) + 4 , 0 }{(s (k)) 2 + 3 s (k) + j, j ,

(s(k)) 2 + 3 s(k) + j, j + 1 ,(s(k)) 2 + 3 s(k) + j, j

1 ,

(s(k)) 2 + 3 s(k) + j, j : 1 j s (k) + 2 }(s(k)) 2 + 5 s(k) + 4 j, j ,(s(k)) 2 + 5 s(k) + 4 j, j + 1 ,(s(k)) 2 + 5 s(k) + 4 j, j 1 ,(s(k)) 2 + 5 s(k) + 4 j, j :

1 j s (k) + 1 }0, otherwise


5/26


Similar representations are possible for the ninth, tenth and eleventh problems. In [28]we introduce eight modications of these problems, giving formulae for their ( k, i )-th membersa k,i .

Essentially more interesting is Problem 103 from [10]:Smarandache numerical carpet:has the general form

1

1 a 1

1 a b a 1

1 a b c b a 1

1 a b c d c b a 1

1 a b c d e d c b a 1

1 a b c d e f e d c b a 1

1 a b c d e f g f e d c b a 1

1 a b c d e f e d c b a 1

1 a b c d e d c b a 1

1 a b c d c b a 1

1 a b c b a 1

1 a b a 1

1 a 1

1

On the border of level 0, the elements are equal to 1;they form a rhomb.

Next, on the border of level 1, the elements are equal to a;where a is the sum of all elements of the previous border;the as form a rhomb too inside the previous one.

Next again, on the border of level 2, the elements are equal to b;


6/26


where b is the sum of all elements of the previous border;the bs form a rhomb too inside the previous one.

And so on ...

The above square, that Smarandache named rhomb , corresponds to the square from ourconstruction for the case of s = 6, if we begin to count from s = 1, instead of s = 0. In [10] aparticular solution of the Problem 103 is given, but there a complete solution is not introduced.We will give a solution below rstly for the case of Problem 103 and then for a more generalcase.

It can be easily seen that the number of the elements of the s-th square side is s + 2 andtherefore the number of the elements from the contour of this square is just equal to 4 s + 4.

The s-th square can be represented as a set of sub-squares, each one included in the next.Let us number them inwards, so that the outmost (boundary) square is the rst one. As it iswritten in Problem 103, all of its elements are equal to 1. Hence, the value of the elements of

the subsequent (second) square will be (using also the notation from problem 103):

a 1 = a = ( s + 2) + ( s + 1) + ( s + 1) + s = 4( s + 1);

the value of the elements of the third square will be

a 2 = b = a (4(s 1) + 4 + 1) = 4( s + 1)(4 s + 1);the value of the elements of the fourth square will be

a 3 = c = b(4( s 2) + 4 + 1) = 4( s + 1)(4 s + 1)(4 s 3);

the value of the elements of the fth square will be

a 4 = d = c(4(s 3) + 4 + 1) = 4( s + 1)(4 s + 1)(4 s 3)(4 s 7);etc.,where the square, corresponding to the initial square (rhomb) , from Problem 103 has theform

1

1 a1 a1 1

1 a1 a2 a2 a1 11 a1 a2 a3 a3 a2 a1 1

1 a1 a2 a2 a1 11 a1 a1 1

1


7/26


It can be proved by induction that the elements of this square that stay on t-th place aregiven by the formula

a t = 4( s + 1)t2

i=0(4s + 1

4i).

If we would like to generalize the above problem, we can construct, e.g., the followingextension:

x

x a 1 a1 x

x a 1 a2 a2 a1 xx a 1 a2 a3 a3 a2 a1 x

x a 1 a2 a2 a1 xx a 1 a1 x

x

where x is given number. Then we obtain

a 1 = 4( s + 1) x

a 2 = 4( s + 1)(4 s + 1) x

a3 = 4( s + 1)(4 s + 1)(4 s 3)xa 4 = 4( s + 1)(4 s + 1)(4 s 3)(4 s 7)xetc. and for t 1

a t = 4( s + 1)t2

i=0(4s + 1 4i)x.

where it assumed that1

i=0 = 1 .

3. The results in this section are taken from [21].The 15-th Smarandaches problem from [10] is the following: Smarandaches simple num-bers:

2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 25, 26, 27,

29, 31, 33, . . .

A number n is called Smarandaches simple number if the product of its proper divisors is less than or equal to n . Generally speaking, n has the form n = p, or n = p2 , or n = p3 , or n = pq , where p and q are distinct primes.


8/26


Let us denote: by S - the sequence of all Smarandaches simple numbers and by sn - then -th term of S ; by P - the sequence of all primes and by pn - the n-th term of P ; by P 2 - thesequence { p2n }n =1 ; by P 3 - the sequence { p3n }n =1 ; by PQ - the sequence { p.q } p,q P , where p < q .

In the present section we nd S (n ) in an explicit form and using this, we nd the n-thterm of S in explicit form, too.

First, we note that instead of P (n ) we use the notation (n ).Hence

P 2 (n ) = ( n ), P 3 (n ) = ( 3 n ),Thus, using the denition of S , we get

S (n ) = (n ) + ( n ) + ( 3 n ) + PQ(n ) (4.1)Our rst aim is to express S (n ) in an explicit form. For (n ) some explicit formulae are

proposed in [18]. Other explicit formulae for (n ) are given in [14]. One of them is known asMinacs formula. It is given below

(n ) =n

k =2

[(k 1)! + 1

k [(k 1)!

k ]]. (4.2)

Therefore, the problem of nding of explicit formulae for functions (n ), ( n ), ( 3 n ) is solvedsuccessfully. It remains only to express PQ(n ) in an explicit form.

Let k {1, 2, . . . , ( n )} be xed. We consider all numbers of the kind pk q , which p P ,q > p k for which pk .q n . The quality of these numbers is ( n pk ) ( pk ), or which is the same

( n pk

) k. (4.3)When k = 1 , 2, . . . , ( n ), the number pk .q , as dened above, describe all numbers of the

kind p.q , with p, q P , p < q, p.q < n . But the quantity of the last numbers is equal to PQ(n ).Hence

PQ(n ) = ( n )

k=1

( ( n pk

) k), (4.4)because of (4.3). The equality (4.4), after a simple computation yields the formula

PQ(n ) = ( n )

k =1

( n pk

) ( n )( ( n ) + 1)

2 . (4.5)

In [20] the identity (b)

k =1

( n pk

) = (nb

). (b) + ( n2 ) ( nb )

k=1

( n

p ( nb )+ k) (4.6)

is proved, under the condition b > 2 (b is a real number). When ( n2 ) = (nb ), the right hand-

side of (4.6) is reduced to ( nb ). (b). In the case b = n and n 4 equality (4.6) yields ( n )

k =1

( n pk

) = ( ( n ))2 + ( n2 ) ( n )

k=1

( n

p ( n )+ k). (4.7)


9/26


If we compare (4.5) with (4.7) we obtain for n 4

PQ(n ) = ( n )( ( n ) 1)

2 +

( n2 ) ( n )

k=1

( n

p ( n )+ k

). (4.8)

Thus, we have two different explicit representations for PQ(n ). These are formulae (4.5)and (4.8). We note that the right hand side of (4.8) reduces to ( n )( ( n )1)2 , when ( n2 ) = ( n ).

Finally, we observe that (4.1) gives an explicit representation for S (n ), since we may useformula (4.2) for (n ) (or other explicit formulae for (n )) and (4.5), or (4.8) for PQ(n ).

The following assertion solves the problem for nding of the explicit representation of sn .Theorem 4.1. The n -th term sn of S admits the following three different explicit representa-tions:

sn =(n )

k =0

[ 1

1 + [ S (n )n ]]; (4.9)

s n = 2(n )

k =0

(2[ S (n )

n ]); (4.10)

s n = (n )

k=0

1(1 [ S (n )n ])

, (4.11)

where(n ) [

n 2 + 3 n + 44

], n = 1 , 2, . . .

We note that (4.9)-(4.11) are representations using, respectively, oor function, Rie-manns Zeta-function and Eulers Gamma-function. Also, we note that in (4.9)-(4.11) S (k) isgiven by (4.1), (k) is given by (4.2) (or by others formulae like (4.2)) and PQ(n ) is given by(4.5), or by (4.8). Therefore, formulae (4.9)-(4.11) are explicit.

4. The results in this section are taken from [6].The 17-th problem from [10] (see also the 22-nd problem from [16]) is the following:Smarandaches digital products:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9

, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

,

0, 2, 4, 6, 8, 10, 12, 14, 16, 18

, 0, 3, 6, 9, 12, 15, 18, 21, 24, 27

,

0, 4, 8, 12, 16, 20, 24, 28, 32, 36

, 0, 5, 10, 15, 20, 25 . . .

( d p(n )is the product of digits.)Let the xed natural number n have the form n = a1a 2 . . . a k , where a1 , a 2 , . . . , a k

{0, 1, . . . , 9} and a1 1. Therefore,

n =k

i=1

a i 10i1 .


10/26


Hence, k = [log10 n ] + 1 and

a 1(n ) a1 = [ n

10k1 ],

a2(n ) a 2 = [n a 110k1

10k2 ],

a3(n ) a 3 = [n a 110k1 a 210k2

10k3 ],

. . .

a [log 10 (n )] (n ) ak1 = [n a 110k1 . . . a k2102

10 ],

a [log 10 (n )]+1 (n ) ak = n a 110k1 . . . a k110.Obviously, k , a 1 , a 2 , . . . , a k are functions only of n . Therefore,

d p(n ) =[log 10 (n )]+1

i =1

a i (n ).

5. The results in this section are taken from [4, 27].The 20-th problem from [10] is the following (see also Problem 25 from [16]):

Smarandache devisor products:

1, 2, 3, 8, 5, 36, 7, 64, 27, 100, 11, 1728, 13, 196, 225, 1024, 17, 5832, 19,

8000, 441, 484, 23, 331776, 125, 676, 729, 21952, 29, 810000, 31, 32768,

1089, 1156, 1225, 10077696, 37, 1444, 1521, 2560000, 41, . . .

( P d (n ) is the product of all positive divisors of n .)The 21-st problem from [10] is the following (see also Problem 26 from [16]):

Smarandache proper devisor products:

1, 1, 1, 2, 1, 6, 1, 8, 3, 10, 1, 144, 1, 14, 15, 64, 1, 324, 1, 400, 21, 22, 1,

13824, 5, 26, 27, 784, 1, 27000, 1, 1024, 33, 34, 35, 279936, 1, 38, 39,

64000, 1, . . .

( pd (n ) is the product of all positive divisors of n but n .)Let us denote by (n ) the number of all devisors of n . It is well-known (see, e.g., [13]) that

P d (n ) = n (n ) (6.1)and of course, we have

pd (n ) = P d (n )

n . (6.2)

But (6.1) is not a good formula for P d (n ), because it depends on function and to express (n ) we need the prime number factorization of n .


11/26


Below, we give other representations of P d (n ) and pd (n ), which do not use the primenumber factorization of n .Proposition 6.1. For n 1 representation

P d (n ) =n

k=1

k [nk ][ n

1k ] (6.3)

holds.Here and further the symbols

k/n and

k/n

mean the product and the sum, respectively, of all divisors of n .Let

(n, k ) [n

k ][n

1

k ]

=1, if k is a divisor of n

0, otherwise

The following assertion is obtained as a corollary of (6.2) and (6.3).Proposition 6.2. For n 1 representation

pd (n ) =n 1

k=1

k [nk ][ n

1k ]

holds.For n = 1 we have

pd (1) = 1 .

Proposition 6.3. For n 1 representation

P d (n ) =n

k =1

[nk ]![n 1k ]!

(6.5)

holds, where here and further we assume that 0! = 1.Now (6.2) and (6.5) yield.


pd (n ) =n

k=2

[nk ]![n 1k ]!

holds.Another type of representation of pd (n ) is the following


pd (n ) =n 2

k=1

(k!)(n,k )(n,k +1) ,


12/26


where (n, k ) is given by (6.4).Further, we need the following

Theorem 6.1. [22] For n 2 the identityn

k =2

[nk

]! =n 1

k=1

(k!)[nk ][ nk +1 ] (6.6)

holds.Now, we shall deduce some formulae for

n

k=1

P d (k) andn

k =1

pd (k).

Proposition 6.6. Let f be an arbitrary arithmetic function. then the identityn

k =1(P d (k)) f (k ) =

n

k=1k(n,k ) (6.7)

holds, where

(n, k ) =[ nk ]

s =1

f (ks ).

Now we need the followingLemma 6.1. For n 1 the identity

n

k=1

[nk

]! =n

k=1

k [nk ]

holds.Proposition 6.7. For n 1 the identity

n

k =1

P d (k) =n

k=1

[nk

]! (6.8)

holds. As a corollary from (6.2) and (6.8), we also obtainProposition 6.8. For n 2 the identity

n

k=1

pd (k) =n

k =2

[nk

]! (6.9)

holds.From (6.6) and (6.9), we obtain

Proposition 6.9. For n 2 the identityn

k=1

pd (k) =n 1

k=1

(k!)[nk ][ nk +1 ] (6.10)

holds.As a corollary from (6.10) we obtain, because of (6.2)


13/26


Proposition 6.10. For n 1 the identityn

k=1

P d (k) =n

k=1

(k!)[nk ][ nk +1 ] (6.10)

holds.Now, we return to (6.7) and suppose that

f (k) > 0 (k = 1 , 2, . . . ).

Then after some simple computations we obtainProposition 6.11. For n 1 representation

P d (k) =n

k =1

k (n,k ) (6.11)

holds, where

(n, k ) =[ nk ]s =1 f (ks )

[ n 1k ]s =1 f (ks )

f (n ) .

For n 2 representation pd (k) =

n 1

k=1

k (n,k ) (6.12)

holds.Note that although f is an arbitrary arithmetic function, the situation with (6.11) and

(6.12) is like the case f (x) 1, because[ n

k]

s =1 f (ks ) [ n 1

k ]

s =1 f (ks )f (n )

= 1, if k is a divisor of n0, otherwise

Finally, we use (6.7) to obtain some new inequalities, involving P d (k) and pd (k) for k =1, 2, . . . , n .

Putting

F (n ) =n

k=1

f (k)

we rewrite (6.7) asn

k =1

(P d (k))f ( k )F ( n ) =

n

k=1

k ([ nk ]s =1 f (ks )) / (F (n )) . (6.13)

Then we use the well-known Jensens inequalityn

k =1

k xk n

k =1

x kk ,

that is valid for arbitrary positive numbers xk , k (k = 1 , 2, . . . , n ) such that

n

k =1

k = 1 ,


14/26


for the case:xk = P d (k),

k

= f (k)F (n )

.

Thus we obtain from (6.13) inequality

n

k =1

f (k).P d (k) (n

k=1

f (k)) .n

k=1

k ([ nk ]s =1 f (ks )) / (

ns =1 f (s )) . (6.14)

If f (x) 1, then (6.14) yields the inequality1n

n

k=1

P d (k) n

k=1

( n k)[ nk ].

If we put in (6.14)

f (k) = g(k)k

for k = 1 , 2, . . . , n , then we obtain

n

k =1

g(k).pd (k) (n

k =1

g(k)k

).n

k=1

( k k)( [nk ]

s =1g ( ks )

s )/ (ns =1

g ( s )s ) . (6.15)

because of (6.2).Let g(x) 1. Then (6.15) yields the very interesting inequality

( 1H n

n

k =1

pd (k)) H n n

k=1

( k k)H [ nk ] ,

where H m denotes the m -th partial sum of the harmonic series, i.e.,

H m = 11

+ 12

+ . . . + 1m

.

All of the above inequalities become equalities if and only if n = 1.

6. The results in this section are taken from [29].The 25-th and the 26-th problems from [10] (see also the 30-th and the 31-st problems

from [16]) are the following:

Smarandaches cube free sieve:

2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 25, 26,

28, 29, 30, 31, 33, 34, 35, 36, 37, 38, 39, 41, 42, 43, 44, 45, 46, 47, 49, 50,

51, 52, 53, 55, 57, 58, 59, 60, 61, 62, 63, 65, 66, 67, 68, 69, 70, 71, 73, . . .

Denition: from the set of natural numbers (except 0 and 1):- take off all multiples of 23 (i.e. 8,16,24,32,40,.. . )


15/26


- take off all multiples of 33

- take off all multiples of 53

. . . and so on (take off all multiples of all cubic primes).

Smarandaches m-power free sieve:Denition: from the set of natural numbers (except 0 and 1) take off all multiples of 2m ,afterwards all multiples of 3m . . . and so on(take off all multiples of all m -power primes, m 2).(One obtains all m -power free numbers.)

Here we introduce the solution for both of these problems.For every natural number m we denote the increasing sequence a(m )1 , a

(m )2 , a

(m )3 , . . . of all

m -power free numbers by m . Then we have

1 2 . . . (m 1) m (m + 1) . . .

Also, for m 2 we havem =

m 1

k=1

(2) k

where(2)k = {x|(x1 , . . . , x k 2)(x = x1 .x 2 . . . x k )}

for each natural number k 1.Let us consider m as an innite sequence for m = 2 , 3, . . . . Then 2 is a subsequence of m .

Therefore, the inequalitya (m )n a (2)n

holds for n = 1 , 2, 3, . . . .Let p1 = 2 , p2 = 3 , p3 = 5 , p4 = 7 , . . . be the sequence of all primes. It is obvious that this

sequence is a subsequence of 2. Hence the inequality

a (2)n pnholds for n = 1 , 2, 3, . . . . But it is well-known that

pn (n ) [n 2 + 3 n + 4

4 ]

(see [12]). Therefore, for any m 2 and n = 1 , 2, 3, . . . we havea (m )n

a (2)n

(n ).

Hence, there exits (n ) such that (n ) (n ) and inequality:a (m )n a (2)n (n ).

holds. In particular, it is possible to use (n ) instead of (n ).In [28] we nd the following explicit formulae for a (m )n when m 2 is xed:

a (m )n = (n )

k =0

[ 1

1 + [ m (k )n ]]; (7.1)


16/26


a (m )n = 2 (n )

k =0

(2[m (k)

n ]); (7.2)

a (m )n = (n )

k=0

1(1 [ m (k )n ])

. (7.3)

Thus, the 26-th Smarandaches problem is solved and for m = 3 the 25-th Smarandachesproblem is solved, too.

The following problems are interesting.Problem 7.1. Does there exist a constant C > 1, such that (n ) C.n ?Problem 7.2. Is C 2?

Below we give the main explicit representation of function m (n ), that takes part in for-mulae (7.1) - (7.3). In this way we nd the main explicit representation for a (m )n , that is basedon formulae (7.1) - (7.3), too.

Theorem 7.1. Function m (n ) allows representationm (n ) = n 1 +

s2 {2,3,..., [ m n ]}(1)(s ) .[

ns m

],

where (s ) denotes the number of all different prime divisors of s.

7. The results in this section are taken from [24].The 28-th problem from [10] (see also the 94-th problem from [16]) is the following:

Smarandache odd sieve:

7, 13, 19, 23, 25, 31, 33, 37, 43, 47, 49, 53, 55, 61, 63, 67, 73, 75, 83,

85, 91, 93, 97, . . .

(All odd numbers that are not equal to the difference of two primes).A sieve is to get this sequence:- subtract 2 from all prime numbers and obtain a temporary sequence;- choose all odd numbers that do not belong to the temporary one.

We nd an explicit form of the n-th term of the above sequence, that will be denoted byC = {C n }n =1 below.

Firstly, we shall note that the above denition of C can be interpreted to the following

equivalent form as follows, having in mind that every odd number is a difference of two primenumbers if and only if it is a difference of a prime number and 2:Smarandaches odd sieve contains exactly these odd numbers that cannot be represented as

a difference of a prime and 2.We rewrite the last denition to the following equivalent form, too:Smarandaches odd sieve contains exactly these odd numbers that are represented as a

difference of a composite odd number and 2.We nd an explicit form of the n -th term of the above sequence, using the third denition

of it. Initially, we use the following two assertions.


17/26


Lemma 8.1. For every natural number n 1, C n +1 is exactly one of the numbers: u C n + 2 , v C n + 4 or w C n + 6.Corollary 8.1. For every natural number n 1:

C n +1 C n + 6 .Corollary 8.2. For every natural number n 1:

C n 6n + 1 . (8.1)Now, we return to the Smarandaches problem.In [18] the following three universal explicit formulae are introduced, using numbers C (k) (k =

0, 1, 2, . . . ), that can be used to represent numbers C n :

C n =

k=0

[ 1

1 + [ C (k )n ]],

C n = 2

k=0

(2[C (k)

n ]),

C n =

k =0

1(1 [ C (k )n ])

.

For the present case, having in mind (8.1), we substitute symbol with 6n + 1 in sumk=0 for C n and we obtain the following sums:

C n =6n +1

k=0

[ 1

1 + [ C (k )n ]], (8.2)

C n = 26n +1

k=0

(2[C (k)

n ]), (8.3)

C n =6n +1

k =0

1(1 [ C (k )n ])

. (8.4)

We must explain why C (n ) (n = 1 , 2, 3, . . . ) is represented in an explicit form. It can bedirectly seen that the number of the odd numbers, that are not bigger than n , is exactly equalto

(n ) = n

[n

2], (8.5)

because the number of the even numbers that are not greater than n is exactly equal to [ n2 ].Let us denote by (n ) the number of all odd numbers not bigger than n , that can be

represented as a difference of two primes. According to the second form of the above givendenition, (n ) coincides with the number of all odd numbers m such that m n and m hasthe form m = p 2, where p is an odd prime number. Therefore, we must study all odd primenumbers, because of the inequality m n . The number of these prime numbers is exactly (n + 2) 1. therefore,

(n ) = (n + 2) 1. (8.6)


18/26


Omitting from the number of all odd numbers that are not greater than n the quantity of those numbers that are a difference of two primes, we nd exactly the quantity of these oddnumbers that are not greater than n and that are not a difference of two prime numbers, i.e.,

C (n ). Therefore, the equalityC (n ) = (n ) (n )

holds and from (8.5) and (8.6) we obtain:

C (n ) = ( n [n2

]) ( (n + 2) 1) = n + 1 [n2

] (n + 2) .But (n +2) can be represented in an explicit form, e.g., by Min acs formula and therefore,

we obtain that the explicit form of C (N ) is

C (N ) = N + 1 [N 2

] N +2

k =2

[(k 1)! + 1

k [(k 1)!

k ]], (8.7)

where N 1 is a xed natural number.It is possible to put [ N +32 ] instead of N + 1 [N 2 ] into (8.7).Now, using each of the formulae (8.2) - (8.4), we obtain C n in an explicit form, using (8.7).It can be checked directly that

C 1 = 7 , C 2 = 13 , C 3 = 19 , C 4 = 23 , C 5 = 25 , C 6 = 31 ,

C 7 = 33 , . . .

andC (0) = C (1) = C (2) = C (3) = C (4) = C (5) = C (6) = 0 .

Therefore from (8.2) - (8.4) we have the following explicit formulae for C n

C n = 7 +6n +1

k=7

[ 1

1 + [ C (k )n ]],

C n = 7 + 26n +1

k =7

(2.[C (k)

n ]),

C n = 7 +6n +1

k =7

1(1 [ C (k )n ])

,

where C (k) is given by (8.7).

8. The results in this section are taken from [7, 26].The 46-th Smarandaches problem from [10] is the following:

Smarandaches prime additive complements;

1, 0, 0, 1, 0, 1, 0, 3, 2, 1, 0, 1, 0, 3, 2, 1, 0, 1, 0, 3, 2, 1, 0, 1, 0, 5, 4, 3, 2, 1,

0, 1, 0, 5, 4, 3, 2, 1, 0, 3, 2, 1, 0, 5, 4, 3, 2, 1, 0 . . .


19/26


(For each n to nd the smallest k such that n + k is prime.)

Remarks: Smarandache asked if it is possible to get as large as we want but nite decreasing

k, k 1, k 2, . . . , 2, 1, 0 (odd k) sequence included in the previous sequence - i.e., for any even integer are there two primes those difference is equal to it? He conjectured the answer is negative.

Obviously, the members of the above sequence are differences between rst prime numberthat is greater or equal to the current natural number n and the same n . It is well-known thatthe number of primes smaller than or equal to n is (n ). Therefore, the prime number smallerthan or equal to n is p (n ) . Hence, the prime number that is greater than or equal to n is thenext prime number, i.e., p (n )+1 . Finally, the n -th member of the above sequence will be equalto

p (n )+1 n, if n is not a prime number0, otherwise

We shall note that in [3] the following new formula pn for every natural number n is given:

pn = (n )

i=0

sg (n (i)) ,

where (n ) = [ n2 +3 n +4

4 ] and

sg (x ) =0, if x 0,1, if x > 0.

Let us denote by an the n-th term of the above sequence. Next, we propose a way for

obtaining an explicit formula for an (n = 1 , 2, 3, . . . ). Extending the below results, we give ananswer to the Smarandaches question from his own remark in [10]. At the end, we proposea generalization of Problem 46 and present a proof of an assertion related to Smarandachesconjecture for Problem 46.Proposition 9.1. an admits the representation

a n = p (n 1)+1 n, (9.1)where n = 1 , 2, 3, . . . , is the prime counting function and pk is the k-th term of prime numbersequence.

It is clear that (9.1) gives an explicit representation for an since several explicit formulae

for (k) and pk are known (see, e.g. [14]).Let us dene

n (m ) = m ! + 2 .

Then all numbers

n (m ), n (m ) + 1 , n (m ) + 2 , . . . , n (m ) + m 2are composite. Hence

a n (m ) m 1.


20/26


This proves the Smarandaches conjecture, since m may grow up to innity. Therefore

{a n }n =1 is unbounded sequence.Now, we shall generalize Problem 46.

Letc c1 , c2 , c3 , . . .

be a strictly increasing sequence of positive integers.Denition. Sequence

b b1 , b2 , b3 , . . .is called c-additive complement of c if and only if bn is the smallest non-negative integer, suchthat n + bn is a term of c.

The following assertion generalizes Proposition 1.Proposition 9.2. bn admits the representation

bn = c c (n 1)+1 n,where n = 1 , 2, 3, . . . , c(n ) is the counting function of c, i.e., c(n ) equals to the quantity of cm , m = 1 , 2, 3, . . . , such that cm n .

Letdn cn +1 cn (n = 1 , 2, 3, . . . ).

The following assertion is related to the Smarandaches conjecture from Problem 46.Proposition 9.3. If {dn }n =1 is unbounded sequence, then {bn }n =1 is unbounded sequence,too.Open Problem. Formulate necessary conditions for the sequence {bn }n =1 to be unbounded.

9. The results in this section are taken from [23].Solving of the Diophantine equation

2x 2 3y2 = 5 (10 .1)i.e.,

2x 2 3y2 5 = 0was put as an open Problem 78 by F. Smarandache in [16]. In [28] this problem is solvedcompletely. Also, we consider here the Diophantine equation

l2 6m 2 = 5,i.e.,

l2 6m 2 + 5 = 0and the Pellian equation

u2 6v2 = 1 ,i.e.,

u2 6v2 1 = 0 .


21/26


In [28] we introduce a generalization of the Smarandaches problem 78 from [16].If we consider the Diophantine equation

2x2

3y2 = p, (10.2)

where p = 2 is a prime number, then using [13], Chapter VII, exercise 2 and the same methodas in the case of (10.1), we obtain the following result.Theorem 10.1. (1) The necessary and sufficient condition for solvability of (10.2) is:

p 5(mod24) or p 23(mod24) (10 .3)(2) if (10.3) is valid, then there exist exactly one solution < x, y > N 2 of (10.2) such that

the inequalities

x < 32 .pand

y < 32 .pholds. Every other solution < x, y > N 2 of (10.2) has the form:

x = l + 3 m

y = l + 2 m,

where < l,m > N 2 is a solution of the Diophantine equationl2 6m 2 = p.

The problem how to solve the Diophantine equation, a special case of which is the aboveone, is considered in Theorem 110 from [13].

10. The results in this section are taken from [9]. In [15, 17] F. Smarandache formulatesthe following four problems:

Problem 1. Let p be an integer 3. Then: p is a prime if and only if

( p 3)! is congruent to p 1

2 (mod p).


( p 4)! is congruent to (1)p3 +1

p + 16 (mod p). (11.1)



22/26


( p 5)! is congruent to rh + r 2 1

24 (mod p), (11.2)

with h =

p

24 and r = p

24.

Problem 4. Let p = ( k 1)!h + 1 be a positive integer k > 5, h natural number. Then: p is a prime if and only if

( p k)! is congruent to (1)t h (mod p). (11.3)with t = h + ph + 1 .

Everywhere above x means the inferior integer part of x , i.e., the smallest integer greaterthan or equal to x.

In [28] we discussed these four problems.

Problem 1. Admits the following representation:Let p 3 be an odd number. Then:

p is a prime if and only if ( p 3)! p 1

2 (mod p).

Different than Smarandaches proof of this assertion is given in [28].Problem 2. Is false, because, for example, if p = 7, then (11.1) obtains the form

6 (1)42(mod7) ,where

6 = (7 4)!and

(1)42 = (1)73 +1

86

,

i.e.,6 2(mod7) ,

which is impossible.Problem 3. Can be modied, having in mind that from r = p 24h it follows:

rh + r 2 1

24 = ( p

24h).h +

p2 48 ph + 24 2h 2 124

= ph 24h 2 + p2 1

24 2 ph + 24 h 2 = p2 1

24 ph,i.e., (11.2) has the form

p is a prime if and only if

( p 5)! is congruent to p2 1

24 (mod p),


23/26


Different than the Smarandaches proof of this assertion is given in [28].Problem 4. Also can be simplied, because

t = h + p

h + 1

= h +(k 1)!h + 1

h + 1= h + ( k 1)! + 1 + 1 = h + ( k 1)! + 2 ,

i.e.,(1)t = ( 1)h ,

because for k > 2: (k 1)! + 2 is an even number. Therefore, (11.3) obtains the form


( p k)! is congruent to (1)h h(mod p),Let us assume that (11.4) is valid. We use again the congruences

( p 1) 1(mod p)( p 2) 2(mod p)

. . .

( p (k 1)) (k 1)(mod p)and obtain the next form of (11.4)


( p 1)! (1)h .(1)k1 .(k 1)!.h (mod p)or


( p 1)! (1)h + k

1

.( p 1)(mod p).But the last congruence is not valid, because, e.g., for k = 5 , h = 3 , p = 73 = (5 1)! + 1 1holds

72! (1)9 .72(mod73) , 2i.e.,

72! 1(mod73) ,1 In [28] there is a misprint: 3! instead of 3.2 In [28] there is a misprint: ( 1) 9 instead of ( 1) 7 .


24/26


while from Wilsons Theorem follows that

72! 1(mod73) .

11. The results in this section are taken from [5].In [17] F. Smarandache discussed the following particular cases of the well-known charac-

teristic functions (see, e.g., [11, 30]).12.1) Prime function: P : N {0, 1}, with

P (n ) =0, if n is a prime

1, otherwise

More generally: P k : N k {0, 1}, where k 2 is an integer, and

P k (n 1 , n 2 , . . . , n k ) = 0, if n 1 , n 2 , . . . , n k are all prime numbers1, otherwise

12.2) Coprime function is dened similarly: C k : N k {0, 1}, where k 2 is an integer,and

C k (n 1 , n 2 , . . . , n k ) =0, if n1 , n 2 , . . . , n k are coprime numbers

1, otherwise

In [28] we formulate and prove four assertions related to these functions.Proposition 12.1. For each k, n 1 , n 2 , . . . , n k natural numbers:

P k(n

1, n

2, . . . , n

k) = 1

k

i=1

(1

P (n

i)) .

Proposition 12.2. For each k, n 1 , n 2 , . . . , n k natural numbers:

C k (n 1 , n 2 , . . . , n k ) = 1 k

i =1

k

j = i+1

(1 C 2(n i , n j )) .

Proposition 12.3. For each natural number n:

C (n )+ P (n ) ( p1 , p2 , . . . , p (n )+ P (n )1 , n ) = P (n ).

Proposition 12.4. For each natural number n:

P (n ) = 1 (n )+ P (n )1

i=1(1 C 2( pi , n )) .

Corollary 12.1. For each natural number k, n 1 , n 2 , . . . , n k :

P k (n 1 , n 2 , . . . , n k ) = 1 k

i =1

(n i )+ P (n i )1

j =1

(1 C 2( pj , n i )) .


25/26


References

[1] K.Atanassov, On some of the Smarandaches problems, American Research Press, Lup-

ton, 1999.[2] K.Atanassov, Remarks on some of the Smarandaches problems. Part 1, SmarandacheNotions Journal, Spring, 12 (2001), 82-98.

[3] K.Atanassov, A new formula for the n-th prime number. Comptes Rendus de lAcademieBulgare des Sciences, 7-8-9 (2001).

[4] K.Atanassov, On the 20-th and the 21-st Smarandaches Problems. Smarandache No-tions Journal, Spring , 1-2-3 (2001), 111-113.

[5] K.Atanassov, On four prime and coprime functions, Smarandache Notions Journal,Spring , 1-2-3 (2001), 122-125.

[6] K.Atanassov, On the 17-th Smarandaches Problem, Smarandache Notions Journal,Spring , 1-2-3 (2002), 124-125.

[7] K.Atanassov, On the 46-th Smarandaches Problem, Smarandache Notions Journal,Spring , 1-2-3 (2002), 126-127.

[8] K.Atanassov, On the second Smarandaches Problem, Notes on Number Theory andDiscrete Mathematics, 3(2003), 46-48.

[9] K.Atanassov, On four Smarandaches Problems, Notes on Number Theory and DiscreteMathematics, 1-6 (2005).

[10] Dumitrescu C., V.Seleacu, Some Sotions and Questions in Number Theory, ErhusUniv. Press, Glendale, 1994.

[11] Grauert H., Lieb I., Fischer W, Differential- und Integralrechnung, Springer-Verlag,Berlin, 1967.

[12] Mitrinovic, D., M. Popadic, Inequalities in Number Theory, Nis, Univ. of Nis, 1978.[13] Nagell T., Introduction to Number Theory, John Wiley Sons, Inc., New York, 1950.[14] Ribenboim P. The New Book of Prime Number Records, Springer, New York, 1995.[15] F.Smarandache, Criteria for a number to be prime, Gazeta Matematica, Vol. XXXVI,

No. 2, 49-52, Bucharest, 1981 (in Romanian).[16] F.Smarandache, Only Problems, Not Solutions!, Xiquan Publ. House, Chicago, 1993.[17] F.Smarandache, Collected Papers, Vol. II, Kishinev University Press, Kishinev, 1997.[18] Vassilev-Missana M., Three formulae for n-th prime and six for n-th term for twin

primes, Notes on Number Theory and Discrete Mathematics, 1(2001), 15-20.[19] Vassilev-Missana M., Some explicit formulae for the composite numbers. Notes on

Number Theory and Discrete Mathematics, 2 (2001).[20] Vassilev-Missana M., On one remarkable identity related to function (x), Notes onNumber Theory and Discrete Mathematics, 4 (2001), 129-136.

[21] Vassilev-Missana M., On 15-th Smarandaches problem. Notes on Number Theory andDiscrete Mathematics, 2 (2003) 42-45.

[22] Vassilev-Missana M., Some representations concerning the product of divisors of n .Notes on Number Theory and Discrete Mathematics, 2 (2004), 54-56.

[23] Vassilev M., Atanassov K., Note on the Diophantine equation 2 x2 3y2 = p, Smaran-dache Notions Journal, 1-2-3 (2000) 64-68.


26/26


[24] Vassilev-Missana M., K.Atanassov, On 28-th Smarandaches Problem. Notes on Num-ber Theory and Discrete Mathematics, 2 (2001), 61-64.

[25] Vassilev-Missana M., K.Atanassov, On ve Smarandaches problems. Notes on Number

Theory and Discrete Mathematics, Vol. 10, 2 (2004), 34-53.[26] Vassilev-Missana, M., K. Atanassov, Remarks on the 46-th Smarandaches Problem.

Notes on Number Theory and Discrete Mathematics, 3 (2004), 84-88.[27] Vassilev-Missana, M., K. Atanassov, On two Smarandaches problems, Notes on Num-

ber Theory and Discrete Mathematics, 4 (2004), 106-112.[28] Vassilev-Missana, M., K. Atanassov, Some Smarandache problems, Hexis, Phoenix,

2004.[29] Vassilev P., Vassilev-Missana M., K.Atanassov, On 25-th and 26-st Smarandaches

Problems. Notes on Number Theory and Discrete Mathematics, 4 (2003), 99-104.[30] Yosida K., Function Analysys, Springer-Verlag, Berlin, 1965.

remarks on some of the smarandache’s problem. part 2

Documents